$r(t) = (3t^3 - 2t, 5t^2)$ What is the speed of $r(t)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\sqrt{81t^4 + 64t^2 + 4}$ (Choice B) B $\sqrt{81t^4 - 36t^2 + 4}$ (Choice C) C $(9t^2 - 2, 10t)$ (Choice D) D $\sqrt{9t^2 - 6t + 2}$
Solution: The speed of a parametric curve is the magnitude of its velocity. If $f(t) = (a(t), b(t))$, then speed is: $\| f'(t) \| = \sqrt{ a'(t)^2 + b'(t)^2 }$ Our position function here is $r(t)$. $\begin{aligned} r'(t) &= (9t^2 - 2, 10t) \\ \\ \text{speed} &= ||r'(t)|| \\ \\ &= \sqrt{81t^4 - 36t^2 + 4 + 100t^2} \\ \\ &= \sqrt{81t^4 + 64t^2 + 4} \end{aligned}$ Therefore, the speed of $r(t)$ is $\sqrt{81t^4 + 64t^2 + 4}$.